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Proof step 9 from proof sharing-com (previous step)

P0(H2 ∪ H0), P1(H1 ∪ H0), H2 ⊥ H0, H2 ⊥ H1, H1 ⊥ H0 ⊦ ∃ H5, H4, H3. H5 ⊥ H3 ∧ H5 ⊥ H4 ∧ H4 ⊥ H3 ∧ H2 ∪ H1 ∪ H0 = H5 ∪ H4 ∪ H3 ∧ P1(H5 ∪ H3) ∧ P0(H4 ∪ H3), H1 ⊥ H0 ∧ H1 ⊥ H2 ∧ H2 ⊥ H0 ∧ H2 ∪ H1 ∪ H0 = H1 ∪ H2 ∪ H0 ∧ P1(H1 ∪ H0) ∧ P0(H2 ∪ H0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
P0(H2 ∪ H0), P1(H1 ∪ H0), H1 ⊥ H0, H2 ⊥ H1, H2 ⊥ H0 ⊦ H2 ∪ H1 ∪ H0 = H1 ∪ H2 ∪ H0, ∃ H5, H4, H3. H5 ⊥ H3 ∧ H5 ⊥ H4 ∧ H4 ⊥ H3 ∧ H2 ∪ H1 ∪ H0 = H5 ∪ H4 ∪ H3 ∧ P1(H5 ∪ H3) ∧ P0(H4 ∪ H3)

Rule argument

Additional information

Used simplifier rules: ⊦ H0 ⊥ H1 → H1 ⊥ H0; (s)